Use the t-distribution to find a confidence interval for a mean
μ given the relevant sample results. Give the best point estimate
for μ, the margin of error, and the confidence interval. Assume the
results come from a random sample from a population that is
approximately normally distributed.
A 90% confidence interval for μ using the sample results x¯=142.0, s=52.8, and n=50
Round your answer for the point estimate to one decimal place, and your answers for the margin of error and the confidence interval to two decimal places.
point estimate =
margin of error =
The 90% confidence interval is to .
Point estimate = sample mean = = 142.0
sample standard deviation = s = 52.8
sample size = n = 50
Degrees of freedom = df = n - 1 = 49
At 90% confidence level the t is ,
= 1 - 90% = 1 - 0.90 = 0.10
/ 2 = 0.10 / 2 = 0.05
t /2,df = t0.05,49 = 1.677
Margin of error = E = t/2,df * (s /n)
= 1.677 * (52.8 / 50)
The 90% confidence interval estimate of the population mean is,
- E < < + E
142.0 - 12.52 < < 142.0 + 12.52
129.48 < < 154.52
point estimate = 142.0
margin of error = 12.52
The 90% confidence interval is 129.48 to 154.52
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